Problem 3

Let

P=A_1A_2\cdots A_k

be a convex polygon in the plane. The vertices

A_1,A_2,\dots,A_k

have integer coordinates and lie on a circle. Let

S

be the area of

P

. An odd positive integer

n

is given such that the square of the length of each side of

P

is an integer divisible by

n

. Prove that

2S

is an integer divisible by

n

Step 2 of 5: An inversive generalized Ptolemy relation

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In plain words

Inverting the polygon at one of its own vertices turns the cyclic polygon into a set of collinear image points, and Ptolemy's inequality on the original circle becomes an exact linear relation among the inverted distances.

\frac{A_1A_2}{OA_1\cdot OA_2}+\frac{A_2A_3}{OA_2\cdot OA_3}+\dots+\frac{A_kA_1}{OA_k\cdot OA_1}=2\cdot\frac{A_1A_2\cdots\widehat{A_i}\cdots}{\cdots}

Detailed analysis

Suppose for contradiction the statement fails for some $(k+1)$ -gon $A_1\cdots A_{k+1}$ (all sides' squared lengths divisible by $p^e$ ) with $k$ minimal among counterexamples, so every diagonal fails to be divisible by $p^e$ in the same squared sense (otherwise a diagonal would split the polygon into two smaller cyclic polygons with all relevant squared lengths divisible by $p^e$ , to which the inductive hypothesis on smaller $k$ would apply). Write $O=A_{k+1}$ and invert the plane at $O$ with radius $1$ . Since inversion sends the circumcircle of the $A_i$ to a line, the images of $A_1,\dots,A_k$ become collinear, and the identity $\frac{A_1A_2}{OA_1\cdot OA_2}+\frac{A_2A_3}{OA_2\cdot OA_3}+\dots+\frac{A_{k-1}A_k}{OA_{k-1}\cdot OA_k}=\frac{A_1A_k}{OA_1\cdot OA_k}$ (a telescoping sum of inverted-distance segments along that line) is exactly a generalized Ptolemy relation for the inverted points.